An $\arctan$ integral

[Tolaso J Kos]

Let $a \in \mathbb{R} \setminus \{0\}$. Prove that:

$$\int_0^\infty\frac{x^{a-1}\arctan x}{(x^a+1)^2}\, {\rm d}x=\frac{\pi}{4|a|}$$

[galactus]

If we use parts on this critter by letting

$u=\tan^{-1}(x), \;\ dv=\frac{x^{a-1}}{(1+x^{a})^{2}}dx$, then the resulting integral is the famous:

$$\frac{1}{|a|}\int_{0}^{\infty}\frac{1}{(x^{2}+1)(x^{a}+1)}dx$$

This integral resolves to $\frac{\pi}{4}$ regardless of the value of 'a'.

It is related to the other famous:

$$\int_{0}^{\frac{\pi}{2}}\frac{1}{\tan^{a}(x)+1}dx=\frac{\pi}{4}$$